A DETAILED KINETIC STUDY
The iodination of propanone
CH3COCH3 (aq) + I2 (aq)  CH3COCH2I (aq) + HI (aq)
is a reaction with interesting kinetics. The reaction is acid-catalysed.
From the equation, one might postulate that the rate equation would
be

d[ I 2 ]
 k[CH3COCH3 ]a [ I 2 ]b
dt
A study of the reaction will involve finding a and b.
First: Find the order with respect to iodine
It is arranged that the propanone concentration is much greater than
the iodine concentration, e.g., [CH3COCH3] = 1.00 mol dm3 and [I2]
= 0.00500 rnol dm3 so that, at the end of run, [CH3COCH3] = 0.995
mol dm3, a decrease of 0.5 %. One can say that [CH3COCH3] is
effectively constant, and

d[ I 2 ]
 k1[ I 2 ]b
dt
so that b can be found.
Solutions of known concentration of (a) propanone, (b) iodine in potassium iodide, and (c) an acid buffer of known pH are prepared
and brought to the required temperature in a thermostat bath. The reaction is started by pipetting volumes of the three solutions into
a flask, and a stop watch is started. After a few minutes, a sample of the reacting mixture is pipetted from the solution into a sodium
hydrogencarbonate solution. This stops the reaction instantly by neutralizing the acid. The time at which the reaction stops is
recorded. The iodine that remains is determined by titration against a standard solution of sodium thiosulphate. The analysis is
repeated at intervals of a few minutes. The volume of thiosulphate required is plotted against the time elapsed since the start of the
reaction. The figure attached shows the plot obtained. It is a straight line: the gradient of the graph does not change as the
concentration of
iodine decreases. This shows that the rate of reaction remains constant as the iodine concentration decreases:

d[ I 2 ]
 Cons tan t
dt
The reaction is zero-order with respect to iodine, and the rate equation becomes

d[ I 2 ]
 k[CH3COCH3 ]a
dt
Second: Find the order with respect to propanone
The procedure is repeated with different concentrations of propanone. It is found that doubling the concentration of propanone
doubles the rate of the reaction: the reaction is first-order with respect to propanone:

d[ I 2 ]
 k[CH3COCH3 ]
dt
Third: Find the effect of acid concentration
A set of runs in buffers with different values of pH and fixed concentrations of propanone and iodine shows that the rate of reaction
is proportional to the hydrogen ion concentration. The rate equation is

d[ I 2 ]
 k[CH3COCH3 ][ H  ]
dt
Interpretation of the rate equation
Iodine must be present for iodination to occur, but its concentration does not appear in the rate equation. To explain this, it is
suggested that the reaction takes place in steps. It is the rate of the slowest step that determines the rate of the overall reaction. If
iodine is involved in a step which is too fast to be rate-determining, it will not appear in the rate equation. The suggested
mechanism for the reaction is
This is the slow, rate-determining step. An enol is formed. Since halogens are electrophiles, iodine reacts rapidly with the C=C
bond through its -electrons:
The intermediate formed has a positive charge on a carbon atom, and fast loses a proton to form the iodoketone: